Two solids `A` and `B` floats in water. It is observed that `A` floats with half of its volume immersed and `B` Floats with `2//3` of its volume immersed. The ratio of densities of `A` and `B` is

Options

A. `4 : 3`
B. `2 : 3`
C. `3 : 4`
D. `1 : 3`

Explanation

To find the ratio of the densities of solids A and B, we can use the principle of buoyancy, which states that the buoyant force acting on a floating object is equal to the weight of the fluid displaced by the object.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- Solid A floats with half of its volume immersed in water.
- Solid B floats with two-thirds of its volume immersed in water.
- We need to find the ratio of the densities of solids A and B.

2. **Setting Up the Equations**:
- Let the density of solid A be \( \rho_A \) and the density of solid B be \( \rho_B \).
- The volume of solid A is \( V \), so the volume submerged is \( \frac{1}{2}V \).
- The volume of solid B is also \( V \), so the volume submerged is \( \frac{2}{3}V \).

3. **Applying the Buoyancy Principle**:
- The weight of solid A is given by:
\[ W_A = \rho_A \cdot V \cdot g \]
- The buoyant force acting on solid A is:
\[ F_{bA} = \rho_{water} \cdot \left(\frac{1}{2}V\right) \cdot g \]
- Setting the weight equal to the buoyant force for solid A:
\[ \rho_A \cdot V \cdot g = \rho_{water} \cdot \left(\frac{1}{2}V\right) \cdot g \]
- Simplifying this gives:
\[ \rho_A = \frac{1}{2} \rho_{water} \]

4. **Repeating for Solid B**:
- The weight of solid B is given by:
\[ W_B = \rho_B \cdot V \cdot g \]
- The buoyant force acting on solid B is:
\[ F_{bB} = \rho_{water} \cdot \left(\frac{2}{3}V\right) \cdot g \]
- Setting the weight equal to the buoyant force for solid B:
\[ \rho_B \cdot V \cdot g = \rho_{water} \cdot \left(\frac{2}{3}V\right) \cdot g \]
- Simplifying this gives:
\[ \rho_B = \frac{2}{3} \rho_{water} \]

5. **Finding the Ratio of Densities**:
- Now we can find the ratio of the densities \( \frac{\rho_A}{\rho_B} \):
\[ \frac{\rho_A}{\rho_B} = \frac{\frac{1}{2} \rho_{water}}{\frac{2}{3} \rho_{water}} = \frac{1}{2} \cdot \frac{3}{2} = \frac{3}{4} \]

### Final Answer:
The ratio of the densities of solids A and B is:
\[ \frac{\rho_A}{\rho_B} = \frac{3}{4} \]